cbse ncert solutions for class 11 chemistry chapter 7 · 2019. 11. 26. ·...
TRANSCRIPT
-
Class–XI–CBSE-Chemistry Equilibrium
1 Practice more on Equilibrium www.embibe.com
CBSE NCERT Solutions for Class 11 Chemistry Chapter 7
Back of Chapter Questions
1. A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
(a) What is the initial effect of the change on vapour pressure?
(b) How do rates of evaporation and condensation change initially?
(c) What happens when equilibrium is restored finally, and what will be the final vapour
pressure?
Solution:
(a) If the volume of the container is suddenly increased, then the vapour pressure
would decrease initially. This is because the amount of vapour remains the same, but the
volume increases suddenly. As a result, the same amount of vapour is distributed in a
larger volume.
(b) Since the temperature is constant, the rate of evaporation also remains constant.
When the volume of the container is increased, the density of the vapour phase decreases.
As a result, the rate of collisions of the vapour particles also decreases. Hence, the rate of
condensation decreases initially.
(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the
rate of condensation. In this case, only the volume changes while the temperature remains
constant. The vapour pressure depends on temperature and not on volume. Hence, the
final vapour pressure will be equal to the original vapour pressure of the system.
2. What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO2] = 0.60M, [O2] = 0.82M and [SO3] = 1.90M?
2SO2(g) + O2(g) ⟶ 2SO2(g)
Solution:
Given: [SO2] = 0.60M,
[O2] = 0.82M
[SO3] = 1.90M
Formula : The equilibrium constant (Kc) = =[𝑝𝑟𝑜𝑑𝑢𝑐𝑡]
[𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡]
-
Class–XI–CBSE-Chemistry Equilibrium
2 Practice more on Equilibrium www.embibe.com
Kc =[SO3]
2
[SO2]2[O2]
=(1.90)2M2
(0.60)2(0.821)M3
= 12.239 M−1
Hence, K for the equilibrium is = 12.239 M−1.
3. At a certain temperature and total pressure of 105Pa, iodine vapour contains 40% by volume of I atoms
I2 (g) ⇌ 2I (g)
Calculate Kp for the equilibrium.
Solution:
Given: total pressure of 105Pa
I atoms =40% by volume
Partial pressure=mole fraction × total pressure
Partial pressure of I atoms (p1) =40
100× ptotal (∴ 𝑣𝑜𝑙𝑢𝑚𝑒 ∝ 𝑚𝑜𝑙𝑒 )
=40
100× 105
= 4 × 104Pa
Partial pressure of I2 molecules,
p𝐼2 =60
100× ptotal
=60
100× 105
= 6 × 104Pa
Formula : Kp =(partial pressure of product)
(partial pressure of reactant)
Kp =(pI)
2
(P12)
-
Class–XI–CBSE-Chemistry Equilibrium
3 Practice more on Equilibrium www.embibe.com
=(4 × 104)2Pa2
6 × 104Pa
= 2.67 × 104Pa
4. Write the expression for the equilibrium constant, Kc for each of the following reactions:
(i) 2NOCl (g) ⇌ 2NO (g) + Cl2 (g)
(ii) 2Cu(NO3)2 (s) ⇌ 2CuO (s) + 4NO2 (g) + O2 (g)
(iii) CH3COOC2H5(aq) + H2O(l) ⇌ CH3COOH (aq) + C2H5OH (aq)
(iv) Fe+3 + (aq) + 3OH− (aq) ⇌ Fe(OH)3 (s)
(v) I2 (s) + 5F2(g) ⇌ 2IF5(g)
Solution:
Formula : The equilibrium constant (Kc) = =[𝑝𝑟𝑜𝑑𝑢𝑐𝑡]
[𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡]
For a heterogeneous equilibrium active mass of pure liquid and solid is taken to be 1
(i) 2NOCl (g) ⇌ 2NO (g) + Cl2 (g)
Kc =[NO (g)]2[Cl2 (g)]
[NOCl (g)]2
(ii) 2Cu(NO3)2 (s) ⇌ 2CuO (s) + 4NO2 (g) + O2 (g)
Kc =[CuO (s)]2[NO2 (g)]
4[O2 (g)]
[Cu(NO3)2 (s)]2
= [NO2(g)]4[O2(g)]
(iii) CH3COOC2H5(aq) + H2O(l) ⇌ CH3COOH (aq) + C2H5OH (aq)
Kc =[CH3COOH (aq)]
2[CH3COOH (aq)]4
[CH3COOC2H5(aq)][H2O(l)]
(iv) Fe+3(aq) + 3OH− (aq) ⇌ Fe(OH)3 (s)
Kc =[Fe(OH)3 (s)]
[Fe+3(aq)][OH− (aq)]3=
1
[Fe+3(aq)][OH− (aq)]3
(v) I2 (s) + 5F2(g) ⇌ 2IF5(g)
Kc =[IF5(g)]
2
[I2 (s)][F2(g)]5 =
[IF5(g)]2
[F2(g)]5
-
Class–XI–CBSE-Chemistry Equilibrium
4 Practice more on Equilibrium www.embibe.com
5. Find out the value of Kc for each of the following equilibria from the value of Kp:
(i) 2NOCl (g) ⇌ 2NO (g) + Cl2 (g); Kp = 1.8 × 10−2 at 500 K
(ii) CaCO3 (s) ⇌ CaO(s) + CO2(g); Kp = 167 at 1073 K
Solution:
Formula : The relation between Kp and Kc is given as:Kp = Kc(RT)Δn
(a) (i) 2NOCl (g) ⇌ 2NO (g) + Cl2 (g); Kp = 1.8 × 10−2 at 500 K
Δn = 3 − 2 = 1
R = 0.0831 barLmol−1K−1
T = 500 K
KP = 1.8 × 10−2
Put all values in the formula of KP = Kc(RT)Δn
⇒ 1.8 × 10−2 = Kc(0.0831 × 500)1
⇒ Kc =1.8 × 10−2
0.0831 × 500
= 4.33 × 10−4 (approximately)
(b) (ii) CaCO3 (s) ⇌ CaO(s) + CO2(g); Kp = 167 at 1073 K
Δn = 2 − 1 = 1
R = 0.0831 barLmol−1K−1 T
T = 1073 K
Kp = 167
Now,put all values in a formula Kp = Kc(RT)Δn
⇒ 167 = Kc(0.0831 × 1073)Δn
⇒ Kc =167
0.0831 × 1073
= 1.87 (approximately)
6. For the following equilibrium, Kc = 6.3 × 1014 at 1000 K
-
Class–XI–CBSE-Chemistry Equilibrium
5 Practice more on Equilibrium www.embibe.com
NO (g) + O3 (g) ⇌ NO2 (g) + O2 (g)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions.
What is Kc, for the reverse reaction?
Solution:
formula :The equilibrium constant (Kc) = =[𝑝𝑟𝑜𝑑𝑢𝑐𝑡]
[𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡]
NO (g) + O3 (g) ⇌ NO2 (g) + O2 (g)
It is given that Kc for the forward reaction is Kc =[NO2 (g)]
2[O2 (g)]
[NO (g)][O3 (g)] =6.3 × 1014
Then, Kc′ for the reverse reaction =
[NO (g)][O3 (g)]
[NO2 (g)]2[O2 (g)]
Kc′ =
1
Kc
=1
6.3 × 1014
= 1.59 × 10−15
7. Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
Solution:
For a pure substance (both solids and liquids),
Active mass of [Pure substance] =Number of moles
Volume=
Mass/molecular moles
Volume=
Mass
Volume×Molecular mass
=Density
Molecular mass
Now, the molecular mass and density (at a particular temperature) of a pure substance is
always fixed and is accounted for in the equilibrium constant. Therefore, the values of pure
substances (solid and liquid) are not mentioned in the equilibrium constant expression.
8. Reaction between N2 and O2 takes place as follows:
-
Class–XI–CBSE-Chemistry Equilibrium
6 Practice more on Equilibrium www.embibe.com
2N2 (g) + O2 (g) ⇌ 2N2O (g)
If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed
to form N2O at a temperature for which Kc = 2.0 × 10−37 determine the composition of the
equilibrium mixture.
Solution:
Given: 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 mol N2 = 0.482
initial mol O2 = and 0.933
the volume of container =10 L
Let the concentration of 𝑁2𝑂 at equilibrium be x.
The given reaction is:
2N2(g) + O2(g) ⇌ 2N2O(g) Initial conc. 0.482 mol 0.933 mol 0 At equilibrium (0.482 − x) mol (1.933 − x) mol x mol
Therefore, at equilibrium, in the 10 L vessel:
[N2] =0.482
10= 0.0482 mol L−1and [O2] =
0.933
10= 0.0933molL−1 [N2O] =
x
10molL−1
Formula: The equilibrium constant (Kc) = =[𝑝𝑟𝑜𝑑𝑢𝑐𝑡]
[𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡]
Kc =[N2O(g)]
2
[N2(g)]2[O2(g)]
⇒ 2.0 × 10−37 =(
x10)
2
(0.0482)2(0.0933)
⇒x2
100= 2.0 × 10−37 × (0.0482)2 × (0.0933)
⇒ x2 = 43.35 × 10−40
⇒ x = 6.6 × 10−20
[N2O] =x
10=
6.6 × 10−20
10= 6.6 × 10−21
-
Class–XI–CBSE-Chemistry Equilibrium
7 Practice more on Equilibrium www.embibe.com
9. Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction is given below:
2NO (g) + Br2 (g) ⇌ 2NOBr (g)
When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at a constant
temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO
and Br2.
Solution:
Given: initial mol of NO =0.087
Initial mol of Br2 =0.0437
mol of NOBr is obtained at equilibrium=0.0518
The given reaction is:
2NO (g)2 mol
+ Br2 (g)1 mol
⇋ 2NOBr (g)2 mol
According to stoichiometry , 2 mol of NOBr are formed from 2 mol of NO and 1 mole of Br2 react
, 2 mol of NOBr is formed
Therefore, 0.0518 mol of NOBr is formed from 0.0518
2 mol of Br, or 0.0259 mol of NO.
The amount of NO and Br present initially is as follows:
[NO] = 0.087 mol [Br2] = 0.0437 mol Therefore, the amount of NO present at equilibrium = 0.087 – 0.0518 = 0.0352 mol And, the amount of Br2 present at equilibrium = 0.0437 – 0.0259 = 0.0178 mol
10. At 450K, Kp = 2.0 × 1010 bar−1 for the given reaction at equilibrium.
2SO2(g) + O2(g) ⇌ 2SO3 (g) What is Kc at this temperature?
Solution:
given :
T = 450 K
Kp = 2.0 × 1010bar−1
We know Δn = 2 − 3 = −1
-
Class–XI–CBSE-Chemistry Equilibrium
8 Practice more on Equilibrium www.embibe.com
R = 0.0831 bar L bar K−1mol−1
Formula : Kp = Kc(RT)Δn
Put all values in the above formula
⇒ 2.0 × 1010bar−1 = Kc(0.0831L bar K−1mol−1 × 450K)−1
⇒ Kc =2.0 × 1010bar−1
(0.0831 L bar K−1 mol−1 × 450K)−1
= (2.0 × 1010bar−1)(0.0831LbarK−1mol−1 × 450K)
= 74.79 × 1010L mol−1
= 7.48 × 1011L mol−1
= 7.48 × 1011M−1
11. A sample of HI(g) is placed in the flask at a pressure of 0.2 atm. At equilibrium, the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium?
2HI (g) ⇌ H2 (g) + I2 (g)
Solution:
Given: pressure = 0.2 atm
equilibrium the partial pressure of HI(g) = 0.04 atm
The initial concentration of HI is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm. Therefore, a decrease in the pressure of HI is 0.2 – 0.04 = 0.16. The given reaction is:
formula for Kp =𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑜𝑓 𝑝𝑒𝑟𝑜𝑑𝑢𝑐𝑡
𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡
Kp =pH2 × pI2
(PHI)2
2HI (g) ⇌ H2 (g) + I2 (g)
Initial conc. 0.2 atm, 0 0 At equilibrium 0.04atm 0.16
2
2.15
2
= 0.08 atm = 0.08atm
-
Class–XI–CBSE-Chemistry Equilibrium
9 Practice more on Equilibrium www.embibe.com
=0.08 × 0.08
(0.04)2
=0.0064
0.0016
= 4.0
Hence, the value of Kp for the given equilibrium is 4.0.
12. A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction
N2 (g) + 3H2 (g) ⇌ 2NH3 (g) is 1.7 × 102. Is the reaction mixture at equilibrium? If not, what
is the direction of the net reaction?
Solution:
Given: mol of N2 = 1.57
mol of H2 = 1.92
mol of NH3=8.13
the volume of vessel =20 Liter
temperature = 500 K
Kc = 1.7 × 102
The given reaction is:
N2 (g) + 3H2 (g) ⇌ 2NH3 (g)
The given concentration of various species is
[N2] =1.57
20mol L−1 [H2] =
1.92
20 mol L−1 [NH3] =
8.13
20 mol L−1
Now, reaction quotient Qc is:
Qc =[NH3]
2
[N2][H2]3
=(
(8.13)20 )
2
(1.5720 ) (
1.9220 )
3
= 2.4 × 103
-
Class–XI–CBSE-Chemistry Equilibrium
10 Practice more on Equilibrium www.embibe.com
Since, Qc ≠ Kc, the reaction mixture is not at equilibrium.
Again, Qc > Kc. Hence, the reaction will proceed in the reverse direction.
13. The equilibrium constant expression for a gas reaction is,
Kc =[NH3]
4[O2]5
[NO]4[H2O]6
Write the balanced chemical equation corresponding to this expression.
Solution:
Kc =𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑙𝑎𝑟 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑡
𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑙𝑎𝑟 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡
The balanced chemical equation corresponding to the given expression can be written as:
4NO (g) + 6H2O(g) ↔ 4NH3(g) + 5O2(g)
14. One mole of H2O and one mole of CO is taken in 10 L vessel and heated to 725 K. At equilibrium, 40% of water (by mass) reacts with CO according to the equation,
H2O (g) + CO(g) ⇌ H2 (g) + CO2 (g)
Calculate the equilibrium constant for the reaction.
Solution:
The given reaction is:
H2O (g) + CO(g) ⇌ H2 (g) + CO2 (g)
Initial conc 1
10M
1
10M 0 0
At
equilibrium
1 − 0.4
10M
1 − 0.4
10M
0.4
10M
0.4
10M
= 0.06 M = 0.06 M = 0.04 M = 0.04 M
Therefore, the equilibrium constant for the reaction =𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑙𝑎𝑟 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑡
𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑙𝑎𝑟 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡
-
Class–XI–CBSE-Chemistry Equilibrium
11 Practice more on Equilibrium www.embibe.com
Kc =[H2][CO2]
[H2O][CO]=
[0.04][0.04]
[0.06][0.06] = 0.444 (approximately)
15. At 700 K, the equilibrium constant for the reaction:
H2 (g) + I2 (g) ⇌ 2HI (g)
is 54.8. If 0.5 mol L−1 of HI(g) is present at equilibrium at 700 K, what are the concentration of
H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium
at 700K?
Solution:
Given: equilibrium constant Kc for the reaction H2 (g) + I2 (g) ⇌ 2HI (g) is 54.8.
temperature =700 K
equilibrium constant Kc for the reaction [H2(g)] [I2(g)]
[HI (g)]2 =54.8
Therefore, at equilibrium, the equilibrium constant Kc′ for the reaction
2HI (g) ⇌ H2 (g) + I2 (g)
Kc′ =
[HI (g)]2
[H2(g)] [I2(g)]=
1
54.8
2HI (g) ⇌ H2 (g) + I2 (g)
0.5 x x
moles of HI(g) is present at equilibrium =0.5 mol L−1
[HI] = 0.5mol L−1 will be
Let the concentrations of hydrogen and iodine at equilibrium be x molL−1
[H2] = [I2] = x mol L−1
KC′ =
x × x
(0.5)2=
1
54.8
⇒ x2 =0.25
54.8
⇒ x = 0.06754
x = 0.068 mol L−1
Hence, at equilibrium, [H2] = x = 0.068 mol L−1 ; [I2] = x = 0.068 mol L
−1
-
Class–XI–CBSE-Chemistry Equilibrium
12 Practice more on Equilibrium www.embibe.com
16. What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?
2ICl (g) ⇌ I2 (g) + Cl2(g); Kc = 0.14
Solution:
The given reaction is:
2ICl (g) ⇌ I2 (g) + Cl2(g)
Initial conc. 0.78 M 0 0
Conc. at equilibrium (0.78 − 2x)M xM xM
Now, we can [I2][Cl2]
[ICl]2= Kc
⇒x × x
(0.78 − 2x)2= 0.14
⇒x2
(0.78 − 2x)2= 0.14
⇒x
0.78 − 2x= 0.374
⇒ x = 0.292 − 0.748x
⇒ 1.748x = 0.292
⇒ x = 0.167
Hence, at equilibrium,
[H2] = xM = 0.167M
[I2] = xM = 0.167M
[HI] = (0.78 − 2x)M = [0.78 − 2 × 0.167]M
= 0.446 M
17. Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration
of C2H6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?
C2H6 (g) ⇌ C2H4 (g) + H2 (g)
-
Class–XI–CBSE-Chemistry Equilibrium
13 Practice more on Equilibrium www.embibe.com
Solution:
Let p is the partial pressure exerted by ethene and hydrogen gas (each) at equilibrium. Now, according to the reaction,
C2H6 (g) ⇌ C2H4 (g) + H2 (g)
Initial conc. 4.0 atm 0 0 At equilibrium 4.0 − p p p
Formula : pC2H4×pH2
pC2H6= Kp
⇒p × p
4.0 − p= 0.04
⇒ p2 = 0.16 − 0.04p
⇒ p2 + 0.04p − 0.16 = 0
Now, p =−0.04±√(0.04)2−4×1×(−0.16)
2×1
=−0.04 ± 0.80
2
=0.76
2 (Taking positive value)
= 0.38
18. Ethyl acetate is formed by the reaction between ethanol and acetic acid, and the equilibrium is represented as:
CH3COOH (l) + C2H5OH (l) ⇌ CH3COOC2H5 (l) + H2O (l)
(i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in
excess and is not a solvent in this reaction)
(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is
0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K,
0.214 mol of ethyl acetate is found after some time. Has equilibrium been reached?
Solution:
(i) water is not in excess and is not a solvent in this reaction so all will be considered in
the equilibrium expression
-
Class–XI–CBSE-Chemistry Equilibrium
14 Practice more on Equilibrium www.embibe.com
Reaction quotient ( Qc) =[CH3COOC2H5][H2O]
[CH3COOH][C2H5OH]
(ii) given : temperature = 293 K, initial mol of acetic acid= 1.00 initial mol of ethanol= 0.18 mol of ethyl acetate=0.171
Let the volume of the reaction mixture be V. Also, here we will consider that water is
a solvent and is present in excess.
The given reaction is:
CH3COOH(l) + C2H5OH(l) ⇌ CH3COOC2H5(l) + H2O(l) Initial conc. 1
VM
0.18
VM
0 0
At equilibrium 1 − 0.171
V
0.18 − 0.171
V
0.171
VM
0.171
VM
=
0.829
VM =
0.009
VM
Therefore, the equilibrium constant for the given reaction is:
Kc =[CH3COOC2H3][H2O]
[CH3COOH][C2H5OH] =
0.171
V×
0.171
V0.829
V×
0.009
V
= 3.919 ≈ 3.92
(i) (ii) temperature = 293 K, Initial mol of acetic acid = 1 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 mol of ethanol = 0. 5
mol of ethyl acetate formed = 0.214
Let the volume of the reaction mixture be V.
CH3COOH(l) + C2H5OH(l) ⇌ CH3COOC2H5(l) + H2O(l) Initial conc. 1.0
VM
0.5
VM
0 0
After some
time
10 − 0.214
V
0.5 − 0.214
V
0.214
VM
0.214
VM
=
0.786
VM =
0.286
VM
Therefore, the reaction quotient is,
Qc =[CH3COOC2H5]
[CH3COOH][C2H5OH]
=
0.214V ×
0.214V
0.786V ×
0.286V
= 0.2037
= 0204 (approximately)
-
Class–XI–CBSE-Chemistry Equilibrium
15 Practice more on Equilibrium www.embibe.com
Since Qc < Kc, equilibrium has not been reached.
19. A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was
attained, concentration of PCl5 was found to be 0.5 × 10−1 mol L−1. If value of Kc is 8.3 × 10
−3,
what are the concentrations of PCl3 and Cl2 at equilibrium?
PCl5 (g) ⇌ PCl3 (g) + Cl2(g)
Solution:
Given : temperature =473
After equilibrium concentration of PCl5 = 0.5 × 10−1 mol L−1
Kc = 8.3 × 10−3
Let the concentrations of both PCl3 and Cl2 at equilibrium be x molL−1. The given reaction
is:
PCl5(g) ↔ PCl3(g) + Cl2(g) At equilibrium
0.5 × 10−1mol L−1 x mol L−1 x mol L−1
It if given that the value of the equilibrium constant, Kc is 8.3 × 10−3.
Now we can write the expression for equilibrium as:
[PCl3][Cl2]
[PCl5]= Kc
⇒x × x
0.5 × 10−1= 8.3 × 10−3
⇒ x2 = 4.15 × 10−4
⇒ x = 2.04 × 10−2
= 0.0204 ≈ 0.02
Therefore, at equilibrium,
[PCl3] = [Cl2] = x = 0.02 mol L−1
20. One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and CO2.
FeO (s) + CO (g) → Fe (s) + CO2 (g); Kp = 0.265 atm at 1050 K
-
Class–XI–CBSE-Chemistry Equilibrium
16 Practice more on Equilibrium www.embibe.com
What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures
are: PCO = 1.4 atm and = 0.80 atm?
Solution:
Given: Kp = 0.265 atm
Temperature = 1050K
initial partial pressures of CO (PCO) = 1.4 atm
initial partial pressures of CO2 (PCO2) = 0.80 atm
For the given reaction,
FeO(s)Initially,
+ CO (g)1.4 atm
⇌ Fe (s) + CO2(g)0.80 atm
Qp =pCO2PCO
=0.80
1.4
= 0.571
It is given that Kp = 0.265.
Since Qc > Kp, the reaction will proceed in the backward direction.
Therefore, we can say that the pressure of CO will increase while the pressure of CO2 will decrease.
Now, let the increase in pressure of CO = decrease in pressure of CO2 be p. Then, we can write,
FeO(s)Initially,
+ CO (g)1.4 atm
↔ Fe (s) + CO2(g)0.80 atm
Kp =pCO2PCO
⇒ 0.265 =0.80 − p
1.4 + p
⇒ 0.371 + 0.265p = 0.80 − p
⇒ 1.265p = 0.429
⇒ p = 0.339 atm
Therefore, equilibrium partial pressure of CO2(PCO2) = 0.80 − 0.339 = 0.461 atm
And, equilibrium partial pressure of CO(PCO) = 1.4 + 0.339 = 1.739 atm
-
Class–XI–CBSE-Chemistry Equilibrium
17 Practice more on Equilibrium www.embibe.com
21. Equilibrium constant, Kc for the reaction
N2 (g) + 3H2 (g) ⇌ 2NH3 (g) at 500 K is 0.061
At a particular time, the analysis shows that composition of the reaction mixture is
3.0 mol L−1 N2, 2.0 mol L−1 H2 and 0.5 mol L
−1 NH3. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?
Solution:
given: 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 = 500 K
Kc =0.061
composition of N2 = 3.0 mol L−1
composition of H2 = 2.0 mol L−1
composition of NH3 = 0.5 mol L−1 .
N2(g) + 3H2(g) ↔ 2NH3(g) At a particular time: 3.0 mol L−1 2.0 mol L−1 0.5 mol L−1
Formula: Qc =[product ]
[𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 ]=
[NH3]2
[N2][H2]3 =
(0.5)2
(3.0)(2.0)3
= 0.0104
It is given that Kc = 0.061
Since Qc ≠ Kc, the reaction is not at equilibrium.
Since Qc < Kc, the reaction will proceed in the forward direction to reach equilibrium.
22. Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:
2BrCl (g) ⇌ Br2 (g) + Cl2 (g) for which Kc = 32 at 500 K. If initially pure BrCl is present at a
concentration of 3.3 × 10−3 mol L−1, what is it's molar concentration in the mixture at equilibrium?
Solution:
Let the amount of bromine and chlorine formed at equilibrium be x. The given reaction is:
-
Class–XI–CBSE-Chemistry Equilibrium
18 Practice more on Equilibrium www.embibe.com
2BrCl(g) ↔ Br2(g) + Cl2(g) Initial conc, 3.3 × 10−3 0 0 At equilibrium 3.3 × 10−3 − 2x x x
Now, we can write,
Kc =[Br2][Cl2]
[BrCl]2
⇒ 32 = x×x
(3.3×10−3−2x)2 by taking root both side
⇒x
3.3 × 10−3 − 2x= 5.66
⇒ x = 18.678 × 10−3 − 11.32x
⇒ 12.32x = 18.678 × 10−3
⇒ x = 1.5 × 10−3
Therefore, at equilibrium,
[BrCl] = (3.3 × 10−3 − 2x) = 3.3 × 10−3 − (2 × 1.5 × 10−3) = 3.3 × 10−3 − 3.0 × 10−3
= 0.3 × 10−3 = 3.0 × 10−4 mol L−1
23. At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with solid carbon has 90.55% CO by mass
C (s) + CO2 (g) ⇌ 2CO (g)
Calculate Kc for this reaction at the above temperature.
Solution:
Given : CO by mass =90.55%
Temperature =1127 k
Pressure = 1 atm
We know 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 CO2 = 44
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 CO = 28
Let the total mass of the gaseous mixture is 100 g. Mass of CO = 90.55 g And, the mass of CO2 = (100 – 90.55) = 9.45 g
-
Class–XI–CBSE-Chemistry Equilibrium
19 Practice more on Equilibrium www.embibe.com
Now, the number of moles of CO (nCO) =weight of CO
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝐶𝑜 =
90.55
28= 3.234 mol
Number of moles of CO2 ( nCO2 ) =weight of CO2
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 CO2 =
9.45
44= 0.215 mol
The partial pressure of CO(pCO) =nCO
nCO+nCO2× ptotal
=3.234
3.234 + 0.215× 1
= 0.938 atm
Partial pressure of CO2 (pCO2) =mole of CO2
total moles × ptotal =
nCO2nCO+nCO2
× ptotal
=0.215
3.234 + 0.215× 1
= 0.062 atm
Therefore, Kp =the partial pressure of the product
the partial pressure of reactant=
[CO]2
[CO2]=
(0.938)2
0.062= 14.19
For the given reaction,
Δn = 2 − 1 = 1
Formula : Kp = Kc(RT)Δn put the values in the formula
⇒ 14.19 = Kc(0.082 × 1127)l
⇒ Kc =14.19
0.082 × 1127
≈ 0.154
24. Calculate a) ∆Ge and b) the equilibrium constant for the formation of NO2 from
NO and O2 at 298K
𝑁𝑂 (𝑔) +1
2 𝑂2 (𝑔) ⇌ 𝑁𝑂2 (𝑔)
where
∆1Go (NO2) = 52.0 kJ/mol
∆Go(NO) = 87.0kJ
mol
∆G𝑜 (O2) = 0 kJ/mol
-
Class–XI–CBSE-Chemistry Equilibrium
20 Practice more on Equilibrium www.embibe.com
Solution:
For the given reaction,
𝑁𝑂 (𝑔) +1
2 𝑂2 (𝑔) ⇌ 𝑁𝑂2 (𝑔)
∆1Go (NO2) = 52.0 kJ/mol
∆Go(NO) = 87.0 kJ/mol
∆Go (O2) = 0 kJ/mol
Formula : ΔGo = ΔGo (products) −ΔGo (Reactants)
ΔGo = 52.0 − {87.0 + 0}
= −35.0 kJ mol−1
(b) formula : ΔGo = −RT ln Kc
ΔGo = −2.303 RT log Kc
Put all values in above formula
log Kc =−35.0×10−3
−2.303×8.314×298
log Kc = 6.134
∴ Kc = antilog (6.134)
Kc = 1.36 × 106
Hence, the equilibrium constant for the given reaction Kc is 1.36 × 106
25. Does the number of moles of reaction products increase, decrease or remain the same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?
(a) PCl5 (g) ⇌ PCl3 (g) + Cl2 (g)
(b) CaO (s) + CO2 (g) ⇌ CaCO3 (s)
(c) 3Fe (s) + 4H2O (g) ⇌ Fe3O4 (s) + 4H2 (g)
Solution:
(a) The number of moles of reaction products will increase. According to Le Chatelier’s
principle, if pressure is decreased, then the equilibrium shifts in the direction in which
the number of moles of gases is more. In the given reaction, the number of moles of
gaseous products is more than that of gaseous reactants. Thus, the reaction will
proceed in the forward direction. As a result, the number of moles of reaction products
-
Class–XI–CBSE-Chemistry Equilibrium
21 Practice more on Equilibrium www.embibe.com
will increase.
(b) The number of moles of reaction products will decrease.
(c) The number of moles of reaction products remains the same.
26. Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction.
(i) COCl2 (g) ⇌ CO (g) + Cl2 (g)
(ii) CH4 (g) + 2S2 (g) ⇌ CS2 (g) + 2H2S (g)
(iii) CO2 (g) + C (s) ⇌ 2CO (g)
(iv) 2H2 (g) + CO (g) ⇌ CH3OH (g)
(v) CaCO3 (s) ⇌ CaO (s) + CO2 (g)
(vi) 4 NH3 (g) + 5O2 (g) ⇌ 4NO (g) + 6H2O(g)
Solution:
(i) COCl2 (g) ⇌ CO (g) + Cl2 (g) ∆𝑛𝑔 = 1
(ii) CH4 (g) + 2S2 (g) ⇌ CS2 (g) + 2H2S (g) ∆𝑛𝑔 = 0
(iii) CO2 (g) + C (s) ⇌ 2CO (g) ∆𝑛𝑔 = 1
(iv) 2H2 (g) + CO (g) ⇌ CH3OH (g) ∆𝑛𝑔 = −2
(v) CaCO3 (s) ⇌ CaO (s) + CO2 (g) ∆𝑛𝑔 = 1
(vi) 4 NH3 (g) + 5O2 (g) ⇌ 4NO (g) + 6H2O(g) ∆𝑛𝑔 = 1
The reaction is given in (ii) ∆𝑛𝑔 = 0 no effect of pressure and no effect on equilibrium because of
the number of moles of gaseous reactants is the same that of gaseous products.
(i), (iii), (iv), (v), and (vi) affected by pressure
The reaction is given in (iv) ∆𝑛𝑔 = −2
will proceed in the forward direction because of the number of moles of gaseous reactants is more
than that of gaseous products.
The reactions are given in (i), (iii), (v), and (vi) in 𝑎𝑙𝑙 ∆𝑛𝑔 = + 1 will shift in the backward
direction because the number of moles of gaseous reactants is less than that of gaseous products.
27. The equilibrium constant for the following reaction is 1.6 × 105 at 1024 K
H2(g) + Br2(g) ⇌ 2HBr(g)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container
at 1024 K.
-
Class–XI–CBSE-Chemistry Equilibrium
22 Practice more on Equilibrium www.embibe.com
Solution:
Kp for the reaction H2(g) + Br2(g) ⇌ 2HBr(g) = 1.6 × 105.
𝐾𝑝= [HBr]2
[H2[Br2]
Therefore, for the reaction 2HBr(g) ⇌ H2(g) + Br2(g), the equilibrium constant will be,
Kp′ =
[H2[Br2]
[HBr]2 =
1
Kp=
1
1.6×105 = 6.25 × 10−6
Now, let p be the pressure of both H2 and Br2 at equilibrium.
Now, we can write,
pH2 × pBr2pHBr
= Kp′
p × p
(10 − 2p)2= 6.25 × 10−6
p
10 − 2p= 2.5 × 10−3
p = 2.5 × 10−2 − (5.0 × 10−3)p
p + (5.0 × 10−3)p = 2.5 × 10−2
(10.5 × 10−3)p = 2.5 × 10−2
p = 2.49 × 10−2 bar ≈ 2.5 × 10−2 bar
Therefore, at equilibrium,
[H2] = [Br2] = p = 2.49 × 10−2 bar
[HBr] = 10 − 2p = 10 − 2 × (2.49 × 10−2) bar
= 9.95 bar ≈ 10 bar
28. Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:
-
Class–XI–CBSE-Chemistry Equilibrium
23 Practice more on Equilibrium www.embibe.com
CH4 (g) + H2O (g) ⇌ CO (g) + 3H2 (g)
(a) Write as the expression for Kp for the above reaction.
(b) How will the values of Kp and composition of equilibrium mixture be affected by
(i) increasing the pressure.
(ii) increasing the temperature.
(iii) using a catalyst?
Solution:
(a) For the given reaction,
Kp =pCO × [PH2
]2
pCH4 × pH2O
(b)
(i)
CH4 (g) + H2O (g) ⇌ CO (g) + 3H2 (g) ; ∆𝑛𝑔 = 2
Kp ∝ 𝑃2
According to Le Chatelier’s principle,on increasing pressure, the equilibrium will shift in the
backward direction.
(ii) According to Le Chatelier’s principle, as the reaction is endothermic, the equilibrium will
shift in the forward direction.
(iii) The equilibrium of the reaction is not affected by the presence of a catalyst. A catalyst only
increases the rate of a reaction. Thus, equilibrium will be attained quickly.
29. Describe the effect of:
(A) addition of H2
(B) addition of CH3OH
(C) removal of CO
(D) removal of CH3OH on the equilibrium of the reaction:
2H2(g) + CO (g) ⇌ CH3OH (g)
Solution:
Given the reaction : 2H2(g) + CO (g) ⇌ CH3OH (g) for this reaction ∆𝑛𝑔 = −2
-
Class–XI–CBSE-Chemistry Equilibrium
24 Practice more on Equilibrium www.embibe.com
(a) According to Le Chatelier’s principle, on the addition of H2(reactant ), the equilibrium of the given reaction will shift in the forward direction to maintain the equilibrium constant reactant has
to decrease .
(b) On addition of CH3OH, the equilibrium will shift in the backward direction to maintain the equilibrium constant reactant has to increase
(c) On removing CO, the equilibrium will shift in the backward direction to maintain the equilibrium constant reactant has to increase
(d) On removing CH3OH, the equilibrium will shift in the forward direction to maintain the
equilibrium constant reactant has to decrease
30. At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride,
PCl5 is 8.3 × 10−3. If decomposition is depicted as,
PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) ∆rHe = 124.0 kJ mol−1
(a) write an expression for Kc for the reaction.
(b) what is the value of Kc for the reverse reaction at the same temperature?
(c) what would be the effect on Kc if
(i) more PCl5 is added
(ii) pressure is increased
(iii) the temperature is increased?
Solution:
(a) for the reaction PCl5 (g) ⇌ PCl3 (g) + Cl2 (g)
Kc =[PCl3(g)][Cl2(g)]
[PCl5(g)]
(a) Value of Kc for the reverse reaction at the same temperature is
PCl3 (g) + Cl2 (g) ⇌ PCl5 (g)
Kc′ =
[PCl5]
[PCl3][Cl2]
Kc′ =
1
Kc=
1
8.3 × 10−3= 1.2048 × 102 = 120.48
(b) (i) Kc =[PCl3(g)][Cl2(g)]
[PCl5(g)]
-
Class–XI–CBSE-Chemistry Equilibrium
25 Practice more on Equilibrium www.embibe.com
more PCl5 is added then 𝐾𝑐 𝑤𝑖𝑙𝑙 𝑛𝑜𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 because of the equilibrium constant does not change by change mole , concentration , pressure. It depends only on temperature and
stoichiometry of the reaction .
(ii) Kc is constant ata constant temperature. Thus, in this case, Kc would not change.
(iii)
PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) ∆rHo = 124.0 kJ mol−1
The reaction is endothermic , the value of Kc increases with an increase in temperature. Since the given reaction in an endothermic reaction, the value of Kc will increase if the temperature is increased.
Or 𝑙𝑛𝑘2
𝑘1= ∆𝐻[
1
𝑇1−
1
𝑇2]
As for endothermic reaction ∆𝐻 = +𝑣𝑒 ; 𝑇2 > 𝑇1; 𝐾2 > 𝐾1
31. Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high-temperature steam. The first stage of two stage reaction involves the formation of CO and
H2. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,
CO (g) + H2O (g) ⇌ CO2 (g) + H2 (g)
If a reaction vessel at 400oC is charged with an equimolar mixture of CO and steam such that
pco = pH2O = 4.0 bar, what will be the partial pressure of H2 at equilibrium? Kp =
10.1 at 400°C
Solution:
Let the partial pressure of both carbon dioxide and hydrogen gas be p. The given reaction is:
It is given that Kp = 10.1
Now,
pCO2 × pH2pCO × pH2O
= Kp
-
Class–XI–CBSE-Chemistry Equilibrium
26 Practice more on Equilibrium www.embibe.com
⇒p×p
(4.0−p)(4.0−p)= 10.1 by talking root both side we get
⇒(p)
(4.0 − p)= 3.178
⇒ p = 12.712 − 3.178p
⇒ 4.178p = 12.712
⇒ p = 3.04
Hence, at equilibrium, the partial pressure of H2 =P= 3.04 bar.
32. Predict which of the following reaction will have appreciable concentration of reactants and products:
a) Cl2 (g) ⇌ 2Cl (g) Kc = 5 × 10−39
b) Cl2 (g) + 2NO (g) ⇌ 2NOCl (g) Kc = 3.7 × 108
c) Cl2 (g) + 2NO2 (g) ⇌ 2NO2Cl (g) Kc = 1.8
Solution:
a) Cl2 (g) ⇌ 2Cl (g) Kc = 5 × 10−39
b) Cl2 (g) + 2NO (g) ⇌ 2NOCl (g) Kc = 3.7 × 108
c) Cl2 (g) + 2NO2 (g) ⇌ 2NO2Cl (g) Kc = 1.8
for (a) the values ofKc = 5 × 10−39 is very less, i.e. reactant is very more
b) Cl2 (g) + 2NO (g) ⇌ 2NOCl (g) Kc = 3.7 × 108 is very high i.e. product is very more
(c) Cl2 (g) + 2NO2 (g) ⇌ 2NO2Cl (g) Kc = 1.8 If the value of Kc lies between 10
−3 and 103, a reaction has appreciable concentration of reactants and products. Thus, the reaction given in (c) will have appreciable concentration
of reactants and products.
33. The value of Kc for the reaction 3O2 (g) ⇌ 2O3 (g) is 2.0 × 10−50 at 25°C. If the equilibrium
concentration of O2 in air at 25°C is 1.6 × 10−2, what is the concentration of O3?
Solution:
Given: Kc = 2.0 × 10−50
-
Class–XI–CBSE-Chemistry Equilibrium
27 Practice more on Equilibrium www.embibe.com
[O2(g)] = 1.6 × 10−2.
The given reaction 3O2(g) ⇌ 2O3(g)
Kc = product of the molar concentration of product
product of the molar concentration of reactant =
[O3(g)]2
[O2(g)]3
Put all the given values in the expression of 𝐾𝑐
2.0 × 10−50 =[O3(g)]
2
[1.6 × 10−2]3
⇒ [O3(g)]2 = 2.0 × 10−50 × (1.6 × 10−2)3
⇒ [O3(g)]2 = 8.192 × 10−56
⇒ [O3(g)]2 = 2.86 × 10−28M
Hence, the concentration of O3 = 2.86 × 10−28M
34. The reaction, CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g) is at equilibrium at 1300 K in a 1L flask. It also contains 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O and an unknown amount of
CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant,
Kc for the reaction at the given temperature is 3.90.
Solution:
Given :Temperature = 1300 K
mol of CO = 0.30
mol of H2 = 0.10
mol of H2O = 0.02
amount of CH4= unknown
Let the concentration of methane at equilibrium be x.
It is given that Kc = 3.90.
-
Class–XI–CBSE-Chemistry Equilibrium
28 Practice more on Equilibrium www.embibe.com
Therefore,
[CH4(g)][H2O(g)]
[CO(g)][H2(g)]3
= Kc
⇒x × 0.02
0.3 × (0.1)3= 3.90
⇒ x =3.90 × 0.3 × (0.1)3
0.02
=0.00117
0.02
= 0.0585 M
= 5.85 × 10−2M
Hence, the concentration of CH4 at equilibrium is 5.85 × 10–2 M.
35. What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species:
HNO2, CN−, HClO4, F
−, OH−, CO32−, and S2−
Solution:
A conjugate acid-base pair is a pair that differs only by one proton.
𝐴𝑐𝑖𝑑 (𝐻𝐴) → 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑎𝑠𝑒(𝐴−) + 𝐻+
𝑏𝑎𝑠𝑒 + 𝐻+ → 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑎𝑐𝑖𝑑
The conjugate acid-base for the given species is mentioned in the table below.
Species Conjugate acid-base
HNO2 NO2−(Conjugate - base)
CN− HCN (Conjugate - acid) HClO4 ClO4
− (Conjugate - base)
F− HF (Conjugate - acid) OH− H2O (Conjugate - acid) OH− O2−(Conjugate - base)
CO32− HCO3
− (Conjugate - acid)
𝑆2− HS− (Conjugate - acid)
-
Class–XI–CBSE-Chemistry Equilibrium
29 Practice more on Equilibrium www.embibe.com
36. Which of the followings are Lewis acids? H2O, BF3, H+, and NH4
+
Solution:
Lewis acids are those species which can accept a pair of electrons. They are electron-deficient
species and having vacant orbital . For example, BF3, H+, And NH4
+ are Lewis acids.
37. What will be the conjugate bases for the Brönsted acids: HF, H2SO4 and HCO3−?
Solution:
𝐴𝑐𝑖𝑑 (𝐻𝐴) → 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑎𝑠𝑒(𝐴−) + 𝐻+
Proton ( 𝐻+) donors are acid
The table below lists the conjugate bases for the given Bronsted acids.
Bronsted acid Conjugate base
HF F− H2SO4 HSO4
−
HCO3− CO3
2−
38. Write the conjugate acids for the following Brönsted bases: NH2−, NH3 and HCOO
−.
Solution:
A conjugate acid of the base formed when base gains proton.
𝑏𝑎𝑠𝑒 + 𝐻+ → 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑎𝑐𝑖𝑑
The table below lists the conjugate acids for the given Bronsted bases.
Species Conjugate acid fo
base
NH2− NH3
NH3 NH4+
HCOO− HCOOH
-
Class–XI–CBSE-Chemistry Equilibrium
30 Practice more on Equilibrium www.embibe.com
39. The species: H2O, HCO3−, HSO4
− and NH3 can act both as Brönsted acids and bases. For each case
give the corresponding conjugate acid and base.
Solution: A conjugate acid-base pair is a pair that differs only by one proton.
𝐴𝑐𝑖𝑑 (𝐻𝐴) → 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑎𝑠𝑒(𝐴−) + 𝐻+
𝑏𝑎𝑠𝑒 + 𝐻+ → 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑎𝑐𝑖𝑑
The table below lists the conjugate acids and conjugate bases for the given species.
Species Conjugate acid Conjugate base
H2O H3O+ OH−
HCO3− H2CO3 CO3
2−
HSO4− H2SO4 SO4
2−
NH3 NH4+ NH2
−
40. Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base:
(a) OH−
(b) F−
(c) H+
(d) BCl3
Solution:
Lewis acids: species which can accept a pair of electrons. They are electron-deficient species
and having vacant orbital
Lewis base: species which can donate a pair of electrons. They are electron-efficient species
(a) OH−is a Lewis base since it can donate its lone pair of electrons. (b) F− is a Lewis base since it can donate a pair of electrons. (c) H+ is a Lewis acid since it can accept a pair of electrons. (d) BCl3 is a Lewis acid since it can accept a pair of electrons.
41. The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10−3 M. what is its pH?
-
Class–XI–CBSE-Chemistry Equilibrium
31 Practice more on Equilibrium www.embibe.com
Solution:
Given,
[H+] = 3.8 × 10−3
∴ pH value of soft drink = − log[H+]
= − log(3.8 × 10−3) (∴ 𝑙𝑜𝑔𝑀 × 𝑁 = 𝑙𝑜𝑔𝑀 + 𝑙𝑜𝑔𝑁)
= −log (3.8 − log103) (∴ 𝑙𝑜𝑔𝑀𝑁 = 𝑁𝑙𝑜𝑔𝑀)
= −log (3.8 + 3)
= −0.58 + 3
= 2.42
42. The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
Solution:
Given, pH = 3.76
It is known that,
pH = − log[H+]
⇒ log[H+] = −pH
⇒ [H+] = antilog(−pH)
= antilog(−3.76)
= 1.74 × 10−4M
Hence, the concentration of hydrogen ion [H+] in the given sample of vinegar is 1.74 × 10−4M.
43. The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10−4, 1.8 × 10−4 and 4.8 ×
10−9 respectively. Calculate the ionization constants of the corresponding conjugate base.
Solution:
Given : The ionization constant of HFat 298K is 6.8 × 10−4
The ionization constant of HCOOH at 298K is 1.8 × 10−4
-
Class–XI–CBSE-Chemistry Equilibrium
32 Practice more on Equilibrium www.embibe.com
The ionization constant of HCN at 298K is 4.8 × 10−9
Formula : KbKa=Kw
Kb =KwKa
Ka of HF = 6.8 × 10−4
Hence, Kb of its conjugate base F− =
Kw
Ka=
10−14
6.8×10−4= 1.5 × 10−11
Ka of HCOOH = 1.8 × 10−4 Hence, Kb of its conjugate base HCOO
− =Kw
Ka=
10−14
1.8×10−4= 5.6 ×
10−11
Ka of HCN = 4.8 × 10−9Hence, Kb of its conjugate base CN
− =Kw
Ka=
10−14
4.8×10−9= 2.08 × 10−6
44. The ionization constant of phenol is 1.0 × 1010. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in
sodium phenolate?
Solution:
Given : Ionization of phenol =1.0 × 1010
Ka =[C6H5O
−][H3O+]
[C6H5OH]
Ka =x×x
0.05−x =1.0 × 1010
As the value of the ionization constant is very less, x will be very small.
We can ignore x in the denominator 0.05 − x ≈ 0.05
x×x
0.05 =1.0 × 1010
∴ x = √1 × 10−10 × 0.05
-
Class–XI–CBSE-Chemistry Equilibrium
33 Practice more on Equilibrium www.embibe.com
= √5 × 10−12
= 2.2 × 10−6M = [H3O+]
Since [H3O+] = [C6H5O
−]
[C6H5O−] = 2.2 × 10−6M
Now, let ∝ be the degree of ionization of phenol in the presence of 0.01 M C6H5ONa.
Also, C6H5OH + H2O ⇌ C6H5O− + 𝐻3
+𝑂
Conc. 0.05 − 0.05α 0.01 + 0.05α 0.05α
[C6H5OH] = 0.05 − 0.05α ≈ 0.05M (∴ 𝛼 ≈ 0)
[C6H5O−] = 0.01 + 0.05α ≈ 0.01M
[H3O+] = 0.05α
Ka =[C6H5O
−][H3O+]
[C6H5OH]
Ka =(0.01)(0.05α)
0.05
1.0 × 10−10 = 0.01α
α = 1 × 10−8
45. The first ionization constant of H2S is 9.1 × 10−8. Calculate the concentration of HS− ion in its
0.1M solution. How will this concentration be affected if the solution is 0.1M in HCl also? If the
second dissociation constant of H2S is 1.2 × 10−13, calculate the concentration of S2− under both
conditions.
Solution:
Given: first ionization constant of H2S = 9.1 × 10−8
the second dissociation constant of H2S is 1.2 × 10−13
(i) To calculate the concentration of HS− ion: Case a: (in the absence of HCl): concentration of H2S solution = 0.1 M
-
Class–XI–CBSE-Chemistry Equilibrium
34 Practice more on Equilibrium www.embibe.com
Let the concentration of HS− = x M.
Then, K𝑎1 =[H+][HS−]
[H2S]
9.1 × 10−8 =(x)(x)
0.1 − x
(9.1 × 10−8)(0.1 − x) = x2
∴ 𝑓𝑖𝑟𝑠𝑡 𝑖𝑜𝑛𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑜𝑓 𝐻2𝑆 = 9.1 × 10−8so x can be ignored and 0.1 − x M ≈ 0.1 M,
we have (9.1 × 10−8)(0.1) = x2.
9.1 × 10−9 = x2
x = √9.1 × 10−9
= 9.54 × 10−5M
⇒ [HS]− = 9.54 × 10−5M
Case b : (in the presence of HCl): concentration of H2S solution = 0.1 M
In the presence of 0.1 M of HCl, let [HS−] be y M
Now, Ka1 =[HS−][H+]
[H2S]
Kaj =[y][0.1 + y]
[0.1 − y]
9.1 × 10−8 =y×0.1
0.1 (∵ 0.1 − y ≈ 0.1M ; 0.1 + y ≈ 0.1M ; )
9.1 × 10−8 = y
⇒ [HS−] = 9.1 × 10−8
(ii) To calculate the concentration of [S2−]
Case c (in the absence of 𝟎. 𝟏 𝐌 𝐇𝐂𝐥):
-
Class–XI–CBSE-Chemistry Equilibrium
35 Practice more on Equilibrium www.embibe.com
HS− ⇌ H+ + S2−
[HS−] = 9.54 × 10−5M (From first ionization, case I)
Let [S2−] be X.
Also, [H+] = 9.54 × 10−5M (From first ionization, case I)
Ka2 =[H+][S2−]
[HS−]
Ka2 =[9.54 × 10−5][X]
[9.54 × 10−5]
1.2 × 10−13 = X = [S2−]
Case d (in the presence of 0.1 M HCl):
Again, let the concentration of HS− be X′ M.
[HS−] = 9.1 × 10−8M (From first ionization, case II)
[H+] = 0.1M (From HCl, case II)
[S−] = X′
HS− ⇌ H+ + S2−
9.1 × 10−8M 0.1 X′
Then, Ka2 =[H+][S2−]
[HS−]
1.2 × 10−13 =(0.1)(X′)
9.1 × 10−8
10.92 × 10−21 = 0.1X′
10.92 × 10−21
0.1= X′
X′ =1.092 × 10−20
0.1
= 1.092 × 10−19M
⇒ Ka1 = 1.74 × 10−5
46. The ionization constant of acetic acid is 1.74 × 10−5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its
pH.
-
Class–XI–CBSE-Chemistry Equilibrium
36 Practice more on Equilibrium www.embibe.com
Solution:
Given: The ionization constant of acetic acid = 1.74 × 10−5
Concentration of acetic acid solution = 0.05 M solution
Method 1
1) CH3COOH ⇌ CH3COO− + H+ Ka = 1.74 × 10
−5
2) H2O + H2O ⇌ H3O− + OH− Kw = 1.0 × 10
−14
Since Ka >> Kw,:
α = √Ka
c= √
1.74×10−5
0.05= 0.018 or 18 % so α can not be neglected so quadratic equation have to
solve
Ka =(0.05α)(0.05α)
(0.05 − 0.05α)
=(0.05α)(0.05α)
0.05(1 − α)
=0.05α2
1 − α
1.74 × 10−5 =0.05α2
1 − α
1.74 × 10−5 − 1.74 × 10−5α = 0.05α2
0.05α2 + 1.74 × 10−5α − 1.74 × 10−5=0
α =−b±√b2−4𝑎𝑐
2a=
=−1.74×10−5±√(1.74×10−5)2−4(0.05)(−1.74×10−5)
2×0.05
α =−1.74×10−5±√(1.74×10−5)2+(0.348 ×10−5)
2×0.05
α =−1.74 × 10−5 ± (1.86 × 10−3)
2 × 0.05
-
Class–XI–CBSE-Chemistry Equilibrium
37 Practice more on Equilibrium www.embibe.com
α =−1.74 × 10−5 + 1.86 × 10−3)
2 × 0.05
α =1.84×10−3
0.1=1.84 × 10−2
[H+] = cα = 0.05 × 1.86 × 10−3
=0.93 × 10−3
1000
= .000093
Method 2
Degree of dissociation,
a = √Kac
c = 0.05 M
Ka = 1.74 × 10−5
Then, α = √1.74×10−5
.05
α = √34.8 × 10−5
α = √3.48 × 10−4
α = 1.86 × 10−2
CH3COOH ⇌ CH3COO− + H+
Thus, concentration of CH3COO− = c. α
= 0.05 × 1.86 × 10−2
= 0.093 × 10−2
= .00093M
Since [OAc−] = [H+],
[H+] = 0.00093 = 0.093 × 10−2
pH = −log[H+]
= − log(0.093 × 10−2)
∴ pH = 3.03
Hence, the concentration of acetate ion in the solution is 0.00093 M and its pH is 3.03.
-
Class–XI–CBSE-Chemistry Equilibrium
38 Practice more on Equilibrium www.embibe.com
47. It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa.
Solution:
given: 𝑝𝐻 = 4.15
Let the organic acid be HA (weak acid)
Concentration of 𝐻𝐴 = 0.01 𝑀
𝑝𝐻 = − log[𝐻+] = 4.15
[𝐻+] = 7.08 × 10−5
𝐾𝑎 =[𝐻+][𝐴]
[𝐻𝐴]
Now, 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑎𝑒𝑘 𝑎𝑐𝑖𝑑 ∶ 𝐻𝐴 ⇌ 𝐻+ + 𝐴−
[𝐻+] = [𝐴−] = 7.08 × 10−5
[𝐻𝐴] = 0.01
Then,
𝐾𝑎 =[𝐻+][𝐴−]
0.01=
(7.08 × 10−5)(7.08 × 10−5)
0.01= 5.01 × 10−7
𝑝𝐾𝑎 = −log𝐾𝑎
= − log(5.01 × 10−7)
𝑝𝐾𝑎 = 6.3001
48. Assuming complete dissociation, calculate the pH of the following solutions:
(A) 0.003 M HCl
(B) 0.005 M NaOH
(C) 0.002 M HBr
(D) 0.002 M KOH
Solution:
(i) 0.003𝑀𝐻𝐶𝑙:
𝐻2𝑂 + 𝐻𝐶𝑙 ⇌ 𝐻3𝑂+ + 𝐶𝑙−
Since HCl is completely ionized,
-
Class–XI–CBSE-Chemistry Equilibrium
39 Practice more on Equilibrium www.embibe.com
[𝐻3𝑂+] = [𝐻𝐶𝑙]
⇒ [𝐻3𝑂+] = 0.003
Now,
𝑝𝐻 = − log[𝐻2𝑂+]
= − log(0.003) = − log(3 × 103) = − [log(3) + 𝑙𝑜𝑔(103)] = −[0.477 + 3] =
= 2.52
Hence, the pH of the solution is 2.52.
(ii) 0.005𝑀𝑁𝑎𝑂𝐻
𝑁𝑎𝑂𝐻(𝑎𝑞) ⇌ 𝑁a+(𝑎𝑞) + HO−(𝑎𝑞)
[𝐻𝑂−] = [𝑁𝑎𝑂𝐻] (∴ 𝑁𝑎𝑂𝐻 𝑖𝑠 𝑠𝑡𝑟𝑜𝑛𝑔 𝑏𝑎𝑠𝑒 )
⇒ [𝐻𝑂−] = 0.005
𝑝𝑂𝐻 = − log[𝐻𝑂−] = − log(0.005)
𝑝𝑂𝐻 = 2.30
∴ 𝑝𝐻 + 𝑝𝑂𝐻 = 𝑝𝐾𝑊 ; 𝑝𝐻 = 14 − 𝑝𝑂𝐻
𝑝𝐻 = 14 − 2.30
= 11.70
Hence, the 𝑝𝐻 of the solution is 11.70.
(iii) 0.002 𝐻𝐵𝑟:
𝐻𝐵𝑟 + 𝐻2𝑂 ⇌ 𝐻3𝑂+ + 𝐵𝑟−
[𝐻3𝑂+] = [𝐻𝐵𝑟]
⇒ [𝐻3𝑂+] = 0.002
∴ 𝑝𝐻 = − log[𝐻3𝑂+]
= − log(0.002)= − log(2 × 103) = −[log(2) + log (103)]= −[log(2) + 3 log (10)]
= 2.69
Hence, the pH of the solution is 2.69.
(iv) 0.002 𝑀 𝐾𝑂𝐻:
𝐾𝑂𝐻(𝑎𝑞) ⇌ 𝐾 +(𝑎𝑞) + 𝑂𝐻
−(𝑎𝑞)
[𝑂𝐻−] = [𝐾𝑂𝐻]
⇒ [𝑂𝐻−] = 0.002
Now, 𝑝𝑂𝐻 = − log[𝑂𝐻−]
= 2.69
-
Class–XI–CBSE-Chemistry Equilibrium
40 Practice more on Equilibrium www.embibe.com
∴ 𝑝𝐻 + 𝑝𝑂𝐻 = 𝑝𝐾𝑊 ; 𝑝𝐻 = 14 − 𝑝𝑂𝐻
𝑝𝐻 = 14 − 2.69
= 11.31
Hence, the pH of the solution is 11.31.
49. Calculate the pH of the following solutions:
(A) 2 g of TlOH dissolved in water to give 2 litre of solution.
(B) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of solution.
(C) 0.3 g of NaOH dissolved in water to give 200 mL of solution.
(D) 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.
Solution:
(a) For 2𝑔 of 𝑇𝑙𝑂𝐻 dissolved in water to give 2 𝐿 of solution:
Moles of 𝑇𝑙𝑂𝐻 =𝑤𝑒𝑖𝑔ℎ𝑡
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 =
2
221
[𝑇𝑙𝑂𝐻(𝑎𝑞)] =
2221
2𝑚𝑜𝑙𝑒/𝐿
=2
2×
1
221𝑀
=1
221𝑀
𝑇𝐼𝑂𝐻(𝑎𝑞) → 𝑇𝐼 +(𝑎𝑞) + 𝑂𝐻
−(𝑎𝑞)
[𝑂𝐻 − (𝑎𝑞)] = [𝑇𝑖𝑂𝐻(𝑎𝑞)] =
1
221𝑀
𝐾𝑤 = [𝐻+][𝑂𝐻−]
10−14 = [𝐻+] (1
221)
221 × 10−14 = [𝐻+]
⇒ 𝑝𝐻 = − log[𝐻+] = −[log(221 × 10−14)] = −[log(2.21 × 10−12)] = −[log(2.21 −12 𝑙𝑜𝑔10 )]
= 11.65
(b) For 0.3 𝑔 of 𝐶𝑎(𝑂𝐻)2 dissolved in water to give 500 mL of solution:
-
Class–XI–CBSE-Chemistry Equilibrium
41 Practice more on Equilibrium www.embibe.com
𝐶𝑎(𝑂𝐻)2 → 𝐶𝑎2+ + 2𝑂𝐻−
Molar mass of 𝐶𝑎(𝑂𝐻)2 = 40 + 2(17) = 74
[𝐶𝑎(𝑂𝐻)2] =0.3
34×
1000
500= 0.0176𝑀
[𝑂𝐻 −(𝑎𝑞)] = 2 × [𝐶𝑎(𝑂𝐻)2] = 2 × 0.0176𝑀
= 0.035 𝑀
[𝐻+] =𝐾𝑤
[𝑂𝐻 − (𝑎𝑞)]
=10−14
0.035𝑀
= 28.33 × 10−14
𝑝𝐻 = −[log(28.33 × 10−14)]
= −[log(0.2833 × 10−12)]
= − [log(0.2833 × 10−12)]
= −[log(0.2833) × (−12𝑙𝑜𝑔10) )
= −(−0.5477 − 12)
= 12.54
(c) For 0.3 𝑔 of 𝑁𝑎𝑂𝐻 dissolved in water to give 200 𝑚𝐿 of solution:
𝑁𝑎𝑂𝐻 → 𝑁𝑎 + (𝑎𝑞) + 𝑂𝐻
−(𝑎𝑞)
Molecular weight of NaOH
[𝑁𝑎𝑂𝐻] =𝑚𝑜𝑙𝑒
𝑣𝑜𝑙𝑢𝑚𝑒(𝑙𝑖𝑡𝑒𝑟)=
0.3
40×
1000
200= 0.0375𝑀
[𝑂𝐻 −(𝑎𝑞)] = 0.0375𝑀
[𝐻+] [𝑂𝐻−] = 𝐾𝑤
[𝐻+] =𝐾𝑤
[𝑂𝐻−]=
𝐾𝑤
[𝑂𝐻−]
[𝐻+] =10−14
0.0375
= 26.66 × 10−14
= 0.2666 × 10−12
-
Class–XI–CBSE-Chemistry Equilibrium
42 Practice more on Equilibrium www.embibe.com
𝑝𝐻 = −[log(0.2666 × 10−12)] = −[log(0.2666) × log (10−12)] = = −[log(0.2666) × (−12log (10 )] =
= 12 −log (0.2666) =12+0.57=12.574
(d) For 1mL of 13.6 𝑀 𝐻𝐶𝑙 diluted with water to give 1 𝐿 of solution:
13.6 × 1 𝑚𝐿 = 𝑀2 × 1000 𝑚𝐿
(Before dilution) (After dilution)
13.6 × 10−3 = 𝑀2 × 1𝐿
𝑀2 = 1.36 × 10−2
[𝐻+] = 1.36 × 10−2
𝑝𝐻 = – 𝑙𝑜𝑔 (1.36 × 10−2
)
= (– 0.1335 + 2)
= 1.866 1.87
50. The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid.
Solution:
Degree of ionization (𝑎) = 0.132 Concentration( 𝑐) = 0.1 𝑀 Thus, the concentration of 𝐻3𝑂
+ = 𝑐. 𝑎 = 0.1 × 0.132
= 0.0132
𝑝𝐻 = − log[𝐻+]
= − log(0.0132)
= 1.879 = 1.88
Now,
𝐾𝑎 = 𝐶𝛼2
= 0.1 × (0.132)2
𝐾𝑎 = 0.00174
𝑝𝐾𝑎 = 2.75
-
Class–XI–CBSE-Chemistry Equilibrium
43 Practice more on Equilibrium www.embibe.com
51. The pH of 0.005M codeine (𝐶18𝐻21𝑁𝑂3) solution is 9.95. Calculate its ionization constant and
pKb.
Solution:
pH of solution = 9.95 Concentration (𝑐) = 0.005 𝑀 Thus, the concentration of 𝐻3𝑂
+ = 𝑐.𝑎
𝑝𝐻 = − log[𝐻+]
9.95 = − log[𝐻+]
[𝐻+ ] = 3.162 × 10−10
𝐾𝑎 = 𝐶𝛼2
𝐶 𝐾𝑎 = 𝐶2𝛼2
𝐾𝑎 =[𝐻+]2
𝐶=
𝑝𝐾𝑎 = 2.75
52. What is the pH of 0.001M aniline solution? The ionization constant of aniline is 4.27 × 10−10 . Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant
of the conjugate acid of aniline.
Solution:
𝐾𝑏 = 4.27 × 10−10
𝑐 = 0.001𝑀
𝛼 =?
𝑘𝑏 = 𝑐𝛼2
4.27 × 10−10 = 0.001 × 𝛼2
4270 × 10−10 = 𝛼2
65.34 × 10−5 = 𝛼 = 6.53 × 10−4
Then, [𝑎𝑛𝑖𝑜𝑛] = 𝑐𝛼 = 0.001 × 65.34 × 10−5
= 0.065 × 10−5
-
Class–XI–CBSE-Chemistry Equilibrium
44 Practice more on Equilibrium www.embibe.com
𝑝𝑂𝐻 = − log(. 065 × 105)
= 6.187
𝑝𝐻 = 7.813
the ionization constant of the conjugate acid of aniline
𝐾𝑎 × 𝐾𝑏 = 𝐾𝑤
∴ 4.27 × 10−10 × 𝐾𝑎 = 𝐾𝑤
𝐾𝑎 =10−14
4.27 × 10−10
= 2.34 × 10−5
Thus, the ionization constant of the conjugate acid of aniline is 2.34 × 10−5.
53. Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains
(a) 0.01M𝑖𝑛 𝐻𝐶𝑙
(b) 0.1M in HCl?
Solution:
𝑐 = 0.05 𝑀
𝑝𝐾𝑎 = 4.74
𝑝𝐾𝑎 = − log(𝐾𝑎)
𝐾𝑎 = 1.82 × 10−5
𝐾𝑎 = 𝑐𝛼2 𝛼 = √
𝐾𝑎𝑐
𝛼 = √1.82 × 10−5
5 × 10−2= 1.908 × 10−2
When 𝐻𝐶𝑙 is added to the solution, the concentration of 𝐻+ ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will
decrease.
Case I: When 0.01 𝑀 𝐻𝐶𝑙 is taken. Let x be the amount of acetic acid dissociated after the addition of 𝐻𝐶𝑙.
-
Class–XI–CBSE-Chemistry Equilibrium
45 Practice more on Equilibrium www.embibe.com
As the dissociation of a very small amount of acetic acid will take place, the values
0.05 – 𝑥 ≈ 0.05
0.01 + 𝑥 ≈ 0.01 taken
𝐾𝑎 =[𝐶𝐻3𝐶𝑂𝑂
−][𝐻+]
[𝐶𝐻3𝐶𝑂𝑂𝐻]
∴ 𝐾𝑎 =(0.01)𝑥
0.05
𝑥 =1.82 × 10−5 × 0.05
0.01
𝑥 = 1.82 × 10−3 × 0.05𝑀
Now,
𝛼 =Amount of acid dissociated
Amount of acid taken intially
=1.82 × 10−3 × 0.05
0.05
= 1.82 × 10−3
Case II: When 0.1 𝑀 𝐻𝐶𝑙 is taken. Let the amount of acetic acid dissociated in this case be 𝑋. As we have done in the first case, the concentrations of various species involved in the reaction are:
[𝐶𝐻3𝐶𝑂𝑂𝐻] = 0.05 − 𝑋 ≈ 0.05𝑀
[𝐶𝐻3𝐶𝑂𝑂−] = 𝑋
[𝐻+] = 0.1 + 𝑋 ≈ 0.1𝑀
𝐾𝑎 =[𝐶𝐻3𝐶𝑂𝑂
−][𝐻+]
[𝐶𝐻3𝐶𝑂𝑂𝐻]
∴ 𝐾𝑎 =(0.1)𝑋
0.05
𝑥 =1.82 × 10−5 × 0.05
0.1
𝑥 = 1.82 × 10−4 × 0.05
Now,
-
Class–XI–CBSE-Chemistry Equilibrium
46 Practice more on Equilibrium www.embibe.com
𝛼 =Amount of acid dissociated
Amount of acid taken initially
=1.82 × 10−4 × 0.05
0.05
= 1.82 × 10−4
54. The ionization constant of dimethylamine is 5.4 × 10−4. Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in
NaOH?
Solution:
Kb = 5,4 × 10−4
c = 0.02 M
Then, α = √Kb
c= √
5.4×10−4
0.02= 0.643
Now, if 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes
complete ionization.
NaOH(aq) ↔ Na+(aq)0.1 M
+ OH−(aq)0.1 M
(𝐶𝐻3)2𝑁𝐻 + 𝐻2𝑂 ⇌ (𝐶𝐻3)2𝑁𝐻2+ + 𝐻𝑂−
At equi. (0.02-x) x x+0.1
Then, [(CH3)2NH2+] = x
(𝐶𝐻3)2𝑁𝐻 = 0.02 − x ≈ 0.02
[OH−] = x + 0.1 ≈ 0.1
⇒ Kb =[(CH3)2NH2
+][OH−]
[(CH3)2NH]
5.4 × 10−4 =x × 0.1
0.02
x = 0.0054
It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.
-
Class–XI–CBSE-Chemistry Equilibrium
47 Practice more on Equilibrium www.embibe.com
55. Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:
(A) Human muscle-fluid, 6.83
(B) Human stomach fluid, 1.2
(c) Human blood, 7.38
(D) Human saliva, 6.4.
Solution:
(a) Human muscle fluid 6.83:
pH = 6.83
pH = − log[H+]
∴ 6.83 = − log[H+]
[H+] = 1.48 × 10−7M
(b) Human stomach fluid, 1.2:
pH = 1.2
pH = − log[H+]
1.2 = − log[H+]
∴ [H+] = 0.063
(c) Human Blood, 7.38
pH = 7.38
pH = − log[H+]
pH = 7.38 = − log[H+]
∴ [H+] = 4.17 × 10−8M
(d) Human saliva, 6.4:
pH = 6.4
pH = − log[H+]
6.4 = −log[H+]
[H+] = 3.98 × 10−7
-
Class–XI–CBSE-Chemistry Equilibrium
48 Practice more on Equilibrium www.embibe.com
56. The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.
Solution:
The hydrogen ion concentration in the given substances ca be calculated by using the given
relation: pH = − log[H+]
(i) the pH of milk = 6.8
Since pH = − log[H+]
6.8 = − log[H+]
[H+] = antilog(−6.8)
[H+] = 1.5 × 19−7M
(ii) the pH of black coffee = 5.0
Since pH = − log[H+]
5.0 = − log[H+]
Log [H+] = −5.0
[H+] = antilog(−5.0) = 10−5M
(iii) the pH of tomato juice = 4.2
Since pH = − log[H+]
4.2 = − log[H+] log
[H+] = antilog(−4.2) = 6.31 × 10−5M
(iv) the pH of lemon juice = 2.2
Since pH = − log[H+]
2.2 = − log[H+] log
[H+] = −2.2
[H+] = antilog(−2.2) = 6.31 × 10−3M
(v) the pH of egg white = 7.8
-
Class–XI–CBSE-Chemistry Equilibrium
49 Practice more on Equilibrium www.embibe.com
Since pH = − log[H+]
7.8 = − log[H+]
Log [H+] = −7.8
[H+] = antilog(−7.8) = 1.58 × 10−8M
57. If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?
Solution:
We know : molecular weight of KOH=39+17=56
moles of KOH=0.561/56
molar concentration of [KOH(aq)] = moles ofsolute
volume of solution(liter) =
0.561
56
(200/1000)L 𝑀 =
0.05 M
KOH(aq) → K+(aq) + OH−(aq)
[OH−] = 0.05 M = [K+]
[H+][OH−] = Kw
[H+] =Kw
[OH−]
[H+] =10−14
0.05= 2 × 10−13M
∴ pH = − log[𝐻+]=
-log(2 × 10−13)
=-[log2 + log10−13
= −[𝑙𝑜𝑔2 − 13log10] = 12.70
-
Class–XI–CBSE-Chemistry Equilibrium
50 Practice more on Equilibrium www.embibe.com
58. The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
Solution:
Solubility of Sr(OH)2 = 19.23 g L⁄
Then, concentration of Sr(OH)2 =19.23
121.63M = 0.1581 M
Sr(OH)2(aq) → Sr2(aq) + 2(OH−)(aq)
∴ [Sr2+] = 0.5181 M
[OH−] = 2 × 0.1581M = 0.3126 M
Now,
Kw = [OH−][H+]
10−14
0.3126= [H+]
⇒ [H+] = 3.2 × 10−14
𝑝𝐻 = − log[𝐻+]
= - [log0.32 + log10−14]
= −[𝑙𝑜𝑔0.32 − 14log10]
pH = 13.495 ≈ 13.50
59. The ionization constant of propanoic acid is 1.32 × 10−5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution
is 0.01M in HCl also?
Solution:
Let the degree of ionization of propanoic acid be α
. Then, representing propionic acid as HA, we have:
𝐻𝐴 + 𝐻2𝑂 ⇌ 𝐻3𝑂+ + 𝐴−
0.05-0.05α ≈ 0.05 0.05α 0.05α
-
Class–XI–CBSE-Chemistry Equilibrium
51 Practice more on Equilibrium www.embibe.com
Ka =[H3O
+][A−]
[HA]=
(0.05α)(0.05α)
0.05= 0.05α2
α = √Ka
0.05= 1.63 × 10−2
Then, [H3O+] = 0.05α = 0.05 × 1.63 × 10−2 = 8.15 × 10−4M
∴ pH = − log[𝐻+]
pH =-[log8.15 × 10−4]
= −[𝑙𝑜𝑔8.15 + 𝑙𝑜𝑔10−4]
= −[𝑙𝑜𝑔8.15 − 4𝑙𝑜𝑔10]
= 4 − 0.911 = 3.09
In the presence of 0.01M of HCl, let α′ is a degree of ionisation.
Then, [H3O+] = 0.01
[A+] = 0.05α′
[HA] = 0.05
Ka =0.01 × 0.05α′
0.05
1.32 × 10−5 = 0.01 × α′
α′ = 1.32 × 10−3
60. The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
Solution:
c = 0.1M
pH = 2.34
− log[H+] = pH
-
Class–XI–CBSE-Chemistry Equilibrium
52 Practice more on Equilibrium www.embibe.com
− log[H+] = 2.34
[H+] = antilog(−2.34)
[H+] = 4.5 × 10−3
Also,
[H+] = cα
4.5 × 10−3 = 0.1 × α
4.5 × 10−3
0.1= α
α = 45 × 10−3 = 0.045
Then,
Ka = cα2
= 0.1 × (45 × 10−3)2
= 202.5 × 10−6
= 2.02 × 10−4
61. The ionization constant of nitrous acid is 4.5 × 10−4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
Solution:
NaNO2 is the salt of a strong base (NaOH) and a weak acid (HNO2).
NO2− + H2O ⇋ HNO2 + OH
−
Kh =[HNO2][OH
−]
[NO2−]
=[HNO2][OH
−][𝐻+]
[NO2−][𝐻+]
∴ 𝑘𝑤 = [OH−][𝐻+] = 10−14 ; 𝐾𝑎 =
[NO2−][𝐻+]
[HNO2 ]
⇒KwKa
=10−14
4.5 × 10−4= 0.22 × 10−10
Now, If x moles of the salt undergoes hydrolysis, then the concentration of various species present in the solution will be:
[NO2−] = 0.04 − x ≈ 0 . 04
[HNO2] = x
[OH−] = x
-
Class–XI–CBSE-Chemistry Equilibrium
53 Practice more on Equilibrium www.embibe.com
Kh =x2
0.04= 0.22 × 10−10
x2 = 0.0088 × 10−10
x = 0.093 × 10−5
∴ [OH−] = 0.093 × 10−5M
[H3O+] =
Kw[HO−]
=10−14
0.093 × 10−5= 10.75 × 10−9M
⇒ pH = − log(10.75 × 10−9)
pH = −[log(10.75 + log (10−9)]
pH = −[log(10.75) − 9log10 )
= 7.96
Therefore, degree of hydrolysis (h)=x
0.04=
0.093×10−5
0.04= 2.325 × 10−5
62. A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
Solution:
pyridinium hydrochloride (𝐶5𝐻7𝐶𝑙𝑁+ ) is salt of a weak base(𝐶5𝐻5𝑁) and strong acid(HCl)
pH = 3.44
We know that,
pH =– log [H+]
-3.44= log [H+]
∴ [H+] = 3.63 × 10−4
𝑝𝑦𝑟𝑖𝑑𝑖𝑛𝑖𝑢𝑚 ℎ𝑦𝑑𝑟𝑜𝑐ℎ𝑙𝑜𝑟𝑖𝑑𝑒 + 𝐻2𝑂 ⇌ 𝑝𝑦𝑟𝑖𝑑𝑖𝑛𝑖𝑢𝑚 + 𝐻𝑂−
Then, Kh =(3.63×10−4)
2
0.02 (∵ concentration = 0.02M)
⇒ Kh = 6.6 × 10−6
-
Class–XI–CBSE-Chemistry Equilibrium
54 Practice more on Equilibrium www.embibe.com
Now, Kh =Kw
Ka
⇒ Ka =KwKh
=10−14
6.6 × 10−6= 1.51 × 10−9
63. Predict if the solutions of the following salts are neutral, acidic or basic: NaCl, KBr, NaCN, NH4NO3, NaNO2 and KF
Solution:
(i) NaCl
2NaCl H O NaOH HCl
Strong base Strong acid
It is a salt of strong acid and strong base .Therefore it is a neutral solution. And its pH=7
(ii) KBr
2KBr H O KOH HBr
Strong base Strong acid
It is a salt of strong acid and strong base .Therefore it is a neutral solution. And its pH=7
(iii) NaCN:
2NaCN H O HCN NaOH
Weak acid Strong base
It is a salt of a weak acid and strong base. Therefore, it is a basic solution.its pH > 7
(iv) NH4NO3
4 3 2 4NH NO H O NH OH HNO
Weak base Strong acid
It is a salt of a strong acid and weak base.Therefore, it is an acidic solution.its pH < 7
(v) NaNO2
2 2 2NaNO H O NaOH HNO
Strong base Weak acid
It is a salt of a weak acid and strong base. Therefore, it is a basic solution.its pH > 7
-
Class–XI–CBSE-Chemistry Equilibrium
55 Practice more on Equilibrium www.embibe.com
(vi) KF
2KF H O KOH HF
Strong base Weak acid
It is a salt of a weak acid and strong base. Therefore, it is a basic solution.its pH > 7
64. The ionization constant of chloroacetic acid is 1.35 × 10−3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?
Solution:
Given : Ka for ClCH2COOH = 1.35 × 10–3.
⇒ Ka = cα2
∴ α = √Kαc
= √1.35 × 10−3
0.1 (∴ concentration of acid = 0.1M)
α = √1.35 × 10−2
= 0.116
∴ [H+] = cα = 0.1 × 0.116
= 0.0116
⇒ pH = − log[H+] = − log[0.0116] ≈ 1.94
ClCH2COONa is the salt of a weak acid ( ClCH2COOH) and a strong base ( NaOH).
ClCH2COO− + H2O ⇌ ClCH2COOH + OH
−
At equi 0.1 x x
Kb =[ClCH2COOH][OH
−]
[ClCH2COO−]
=[ClCH2COOH][OH
−][H+]
[ClCH2COO−][H+]
(∴ [OH−][H+] = 𝐾𝑤 ; 𝐾𝑎 =[ClCH2COO
−][H+]
[ClCH2COOH])
Kh =KwKa
Kh =10−14
1.35 × 10−3
-
Class–XI–CBSE-Chemistry Equilibrium
56 Practice more on Equilibrium www.embibe.com
= 0.740 × 10−11
Also, Kh =x2
0.1 (where x is the concentration of OH− and ClCH2COOH)
0.740 × 10−11 =x2
0.1
0.074 × 10−11 = x2
⇒ x2 = 0.74 × 10−12
x = 0.86 × 10−6
[OH−] = 0.86 × 10−6
∴ [H+] =Kw
0.86 × 10−6
=10−14
0.86 × 10−6
[H+] = 1.162 × 10−8
pH = − log[H+]
=-[log(1.162 × 10−8)]
=-[log1.162-(8log10)]
=log1.162+8log10
= 7.94
65. Ionic product of water at 310 K is 2.7 × 10−14. What is the pH of neutral water at this temperature?
Solution:
Ionic product,
Kw = [H+][OH−]
Let [H+] = x
Since [H+] = [OH−], Kw = x2.
⇒ Kw at 310K is 2.7 × 10−14
∴ 2.7 × 10−14 = x2
-
Class–XI–CBSE-Chemistry Equilibrium
57 Practice more on Equilibrium www.embibe.com
⇒ x = 1.64 × 10−7
⇒ [H+] = 1.64 × 10−7
⇒ pH = − log[H+]
= − log[1.64 × 10−7]
= −[log[1.64 + log10−7]
= -log1.64+7log10
= 6.78
Hence, the pH of neutral water is 6.78.
66. Calculate the pH of the resultant mixtures:
(a) 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl
(b) 10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2
(c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH
Solution:
(a) Moles of H3O+ = molarity × volume (litre) =
25×0.1
1000= 0.0025 mol
Moles of OH− = molarity × volume (litre) =10×0.2×2
1000= 0.0040 mol
Total volume= 10+25
1000𝑙𝑖𝑡𝑒𝑟 = 35 × 10−3 𝑙𝑖𝑡𝑒𝑟
Thus, [OH−] > [H+]
[OH−] = Moles of OH− − Moles of H3O
+
total volume =
0.0040 − 0.1125
35 × 10−3
=0.0015
35 × 10−3mol L⁄ = 0.0428
pOH = − log[ OH− ] = - log(0.0428)
= 1.36
pH = PKw − pOH = 14 − 1.36
= 12.63 (not matched)
-
Class–XI–CBSE-Chemistry Equilibrium
58 Practice more on Equilibrium www.embibe.com
(b) Moles of H3O+ =
2×10×0.01
1000= 0.0002 mol
Moles of OH− =2×10×.01
1000= 0.0002 mol
Since there is neither an excess of H3O+ or OH− so the pH of the solution =7
(c) Moles of H3O+ =
2×10×0.1
1000= 0.002 mol
Moles of OH− =10×0.1
1000= 0.001 mol
Total volume= 10+10
1000𝑙𝑖𝑡𝑒𝑟 = 20 × 10−3 𝑙𝑖𝑡𝑒𝑟
H3O+ > OH−
[H3O+] =
Moles of (OH−) − Moles of (H3O+)
total volume =
0.002 − 0.001
20 × 10−3=
0.001
20 × 10−3=
10−3
20 × 10−3
= 0.05
∴ pH = − log(0.05)
= 1.30
The solution is acidic.
67. Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride
and mercurous iodide at 298K from their solubility product constant are 1.1 × 10−12, 1.2 ×
10−10, 1.0 × 10−38 ,1.6 × 10−5 ,4.5 × 10−29 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦 . Determine also the molarities of individual ions.
Solution:
(1) Silver chromate:
Ag2CrO4 → 2Ag
+ + CrO42−
Then,
Ksp = [Ag+]2[CrO4
2−]
Let the solubility of Ag2CrO4 be s.
⇒ [Ag+] = 2s and [CrO42−] = s
Then,
Ksp = (2s)2 × s = 4s3
-
Class–XI–CBSE-Chemistry Equilibrium
59 Practice more on Equilibrium www.embibe.com
⇒ 1.1 × 10−12 = 4s3
. 275 × 10−12 = s3
s = 0.65 × 10−4M
Molarity of Ag+ = 2s = 2 × 0.65 × 10−4 = 1.30 × 10−4M
Molarity of CrO42− = s = 0.65 × 10−4M
(2) Barium chromate:
BaCrO4 → Ba2+ + CrO4
2−
Ksp = [Ba2+][CrO4
2−]
Let solubility of BaCrO4 be s.
So, [Ba2+] = s and [CrO42−] = s ⇒ KSP = s
2
⇒ 1.2 × 10−10 = s2
⇒ s = 1.09 × 10−5M
Molarity of Ba+ = Molarity of CrO42− = s = 1.09 × 10−5M
(3) Ferric hydroxide:
Fe(OH)3→Fe3+ + 3OH−
Ksp = [Fe3+][OH−]3
Let s be the solubility of Fe(OH)3.
Thus,[Fe3+] = s and [OH−] = 3s
⇒ KSP = s × (3s)3
= s × 27s3
KSP = 27s4
1.0 × 10−38 = 27s4
1.0 × 10−38 = 27s4
0.037 × 10−38 = s4
0. 00037 × 10−36 = s4 ⇒ 1.39 × 10−10M = S
Molarity of Fe3+ = s = 1.39 × 10−10M
Molarity of OH− = 3s = 4.17 × 10−10M
-
Class–XI–CBSE-Chemistry Equilibrium
60 Practice more on Equilibrium www.embibe.com
(4) Lead chloride:
PbCl2 → Pb2+ + 2Cl−
KSP = [Pb2+][Cl−]2
Let KSP be the solubility of PbCl2.
[PB2+] = s and [Cl−] = 2s
Thus, Kxp = s × (2s)2
= 4s3
⇒ 1.6 × 10−5 = 4s3
⇒ 0.4 × 10−5 = s3
4 × 10−6 = s3 ⇒ 1.58 × 10−2M = S. 1
Molarity of Pb2+ = s = 1.58 × 10−2M
Molarity of chloride 2s = 3.16 × 10−2M
(5) Mercurous iodide:
Hg2
I2→Hg22+ + 2I−
Ksp = [Hg22+]
[I−]2
Let s be the solubility of Hg2I2.
⇒ [Hg22+] = s and [I−] = 2s
Thus, Hg2
I2 = s(2s)2 ⇒ Kxy = 4s
3
4.5 × 10−29 = 4s3
1.125 × 10−29 = s3
⇒ s = 2.24 × 10−10M
Molarity of Hg22+ = s = 2.24 × 10−10M
Molarity of I− = 2s = 4.48 × 10−10M
68. The solubility product constant of Ag2CrO4 and AgBr are 1.1 × 10−12 and
5.0×10−13 respectively. Calculate the ratio of the molarities of their saturated solutions.
-
Class–XI–CBSE-Chemistry Equilibrium
61 Practice more on Equilibrium www.embibe.com
Solution:
Let s be th